But L = £ .>' ^ Appendix IL . ._»_ IX I •ff and the polar distance is fixed. Taking base D M, summate ABC from pole o, obtaining D E F M, shewing total load on the left of any vertical. Next make D H = M F, and with H as pole summate D E F to base D j, obtaining D G. A vertical through G will pass through the centre-of-gravity of the load. Set off a c^ A c, ag^ A G, and draws£- K horizontally. Then D L and F K are right and left reactions respectively. The deduction of area DLKM from D E F M gives the two triangular areas, whose verticals indicate shear, or total load on left of section less left reaction.